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I am trying to work out a closed form for the integral

\begin{equation} \int_{0}^{1} \frac{1}{\sqrt{s(1-s)}} \exp\left(-\left(\frac{a}{s} + \frac{b}{1-s}\right) \right) \,ds \end{equation} where $a,b>0$. I tried the susbtitution $\sigma = 1/s$ but did not get very far. I also suspect that some complex variable method could be used. Mathematica cannot solve it, it says it diverges (but clearly it does not), and is even able to compute it numerically without any problem when values are given to $a$ and $b$.

Michael Hardy
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Rogelio Molina
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    I would bet that the answer is heavily related with Bessel functions, since: $$\mathcal{L}(e^{-1/t})=\frac{2}{\sqrt{s}},K_1(2\sqrt{s}),\qquad \mathcal{L}^{-1}(\frac{1}{\sqrt{s(1-s)}})=-i e^{t/2},I_0(t/2).$$ – Jack D'Aurizio Feb 15 '15 at 00:31
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    Letting $x=\sin^2t$ or $x=\cos^2t,~$ for $a=b$ we have $I=2\displaystyle\int_0^\tfrac\pi2e^{\large-4a~\csc^2(2t)}~dt~=~\pi~\text{erfc} \big(2\sqrt a\big)$. – Lucian Feb 15 '15 at 02:08

1 Answers1

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This is a convolution between two inverse Laplace transforms, which is simply the inverse Laplace transform of the product of the LTs. To clarify,

$$f_a(t) = t^{-1/2} e^{-a/t} \implies F_a(s) = \sqrt{\frac{\pi}{s}} e^{-2 \sqrt{a s}} $$

The convolution theorem states that

$$f_a(t) * f_b(t) = \int_0^t dt' f_a(t') f_b(t-t') = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F_a(s) F_b(s) e^{s t}$$

In this case, the convolution integral above is

$$ \frac1{i 2 \pi} \pi \int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-2 (\sqrt{a}+\sqrt{b}) \sqrt{s}} e^{s t} $$

This is an inverse LT derived in this answer. The result is, for $t=1$,

$$\int_0^1 dt' \frac1{\sqrt{t' (1-t')}} \exp{\left (-\frac{a}{t'} - \frac{b}{1-t'} \right )} = \pi \operatorname{erfc}{\left (\sqrt{a} + \sqrt{b} \right )} $$

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Ron Gordon
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